We know that a polynomial function is everywhere continuous.
Therefore, the functions x2+4andx+2 are everywhere continuous.
So, the product function x2+4x+2 is everywhere continuous.
Thus, f(x) is continuous at every x≠2.

At x = 2, we have

(LHL at x = 2) = limx→2-fx=limh→0f2-h=limh→02-h2+42-h+2=84=32

(RHL at x = 2) = limx→2+fx=limh→0f2+h=limh→02+h2+42+h+2=84=32

Also, f2=16

∴ limx→2-fx=limx→2+fx≠f2

Thus, fx is discontinuous at x = 2.

Hence, the only point of discontinuity for fx is x = 2.

(iii)

When x < 0, thenfx=sinxx

We know that sin xas well as the identity function x are everywhere continuous.
So, the quotient function sinxx is continuous at each x < 0.

When x > 0, thenfx=2x+3, which is a polynomial function.
Therefore, fx is continuous at each x > 0.

Now,
Let us consider the point x = 0.
Given: fx=sinxx,ifx<02x+3,ifx≥0

We have
(LHL at x = 0) = limx→0-fx=limh→0f0-h=limh→0f-h=limh→0sin-h-h=limh→0sinhh=1

(RHL at x = 0) = limx→0+fx=limh→0f0+h=limh→0fh=limh→02h+3=3

∴ limx→0-fx≠limx→0+fx

Thus, fx is discontinuous at x = 0.

Hence, the only point of discontinuity for fx is x = 0.

(iv)

When x≠ 0, thenfx=sin3xx

We know that sin 3xas well as the identity function x are everywhere continuous.
So, the quotient function sin3xx is continuous at each x ≠ 0.

We know that sin xas well as the identity function x both are everywhere continuous.
So, the quotient function sinxx is continuous at each x ≠ 0.
Also, cos xis everywhere continuous.
Therefore, sinxx+cosx is continuous at each x ≠ 0.

Let us consider the point x = 0.
Given: fx=sinxx+cosx,ifx≠05,ifx=0

We have
(LHL at x = 0) = limx→0-fx=limh→0f0-h=limh→0f-h=limh→0sin-h-h+cos-h=limh→0sin-h-h+limh→0cos-h=1+1=2

Page No 9.43:

Question 6:

If fx=1-sinxπ-2x2.logsinxlog1+π2-4πx+4x2,x≠π2k,x=π2
is continuous at x = π/2, then k =
(a) -116

(b) -132

(c) -164

(d) -128

Answer:

c-164

If fx is continuous at x=π2, thenlimx→π2fx=fπ2Ifπ2-x=t,then⇒limt→0fπ2-t=fπ2⇒limt→01-sinπ2-t4t2×logsinπ2-tlog1+π2-4ππ2-t+4π2-t2=k⇒limt→01-cost4t2×logcostlog1+π2-2π2+4πt+4π24+t2-πt=k⇒limt→01-cost4t2×logcostlog1-π2+4πt+π2+4t2-4πt=k⇒limt→01-cost4t2×logcostlog1+4t2=k⇒limt→02sin2t216×t24×logcostlog1+4t2=k⇒216limt→0sin2t2t24×logcost4t2log1+4t24t2=k⇒18limt→0sin2t2t22×logcost4t2log1+4t24t2=k⇒18limt→0sin2t2t22×log1-sin2t4t2log1+4t24t2=k⇒18limt→0sin2t2t22×log1-sin2t8t2log1+4t24t2=k⇒164limt→0sin2t2t22×log1-sin2tt2log1+4t24t2=k⇒164limt→0sint2t22×limt→0log1-sin2tt2limt→0log1+4t24t2=k⇒1641×limt→0-sin2tlog1-sin2tt2-sin2t=k⇒-164limt→0sin2tlog1-sin2tt2-sin2t=k⇒-164limt→0sintt2limt→0log1-sin2t-sin2t=k⇒-164limt→0sintt2limt→0log1-sin2t-sin2t=k⇒k=-164∵limx→0log1-xx=1

If fx is continuous for all x, then it will be continuous at x = 0 as well.

So, if fx is continuous at x = 0, thenlimx→0fx=f0

⇒limx→0-aa+x+a-xa2-ax+x2+a2+ax+x2=f0⇒-2aaa2+a2=f0⇒-2aaa+a=f0⇒f0=-a

Page No 9.44:

Question 17:

The functionfx=1,x≥11n2,1n<x<1n-1,n=2,3,...0,x=0
(a) is discontinuous at finitely many points
(b) is continuous everywhere
(c) is discontinuous only at x=±1n, n ∈ Z − {0} and x = 0
(d) none of these

Answer:

Given: fx=1,x≥11n2,1n<x<1n-10,x=0

⇒fx=1,-1≤x≤11n2,1n<x<1n-10,x=0

Case 1: x>1orx<-1andx>1

Here,fx=1, which is the constant function
So, fx is continuous for all x≥1orx≤-1andx≥1.

Case 2: 1n<x<1n-1,n=2,3,4,...

Here,fx=1n2,n=2,3,4,...fx=1n2,n=2,3,4,..., which is also a constant function.

So, fx is continuous for all 1n<x<1n-1,n=2,3,4,....

Case 3: Consider the points x = -1 and x = 1.

We haveLHLatx=-1=limx→-1-fx=limx→-1-1=1RHLatx=-1=limx→-1+fx=limx→-1+14=14∵fx=14for-1<x<12,whenn=2Clearly,limx→-1-fx≠limx→-1+fxatx=-1So,fxisdiscontinuousatx=-1.

Similarly, f(x) is discontinuous at x = 1.

Case 4: Consider the point x = 0.

We havelimx→0-fx=limh→0f1n-h=limh→0f1n-h=1n-12

limx→0+fx=limh→0f1n+h=limh→0f1n+h=1n2

limx→0+fx≠limx→0-fx

Thus, fx is discontinuous at x=0.

At x = 0, we havelimx→0-fx≠0=f0

So, fx is discontinuous at x=0.

Case 5: Consider the point x=1n,n=2,3,4,...

We havelimx→1n-fx=limh→0f1n-h=limh→0f1n-h=1n-12

limx→1n+fx=limh→0f1n+h=limh→0f1n+h=1n2

limx→1n+fx≠limx→1n-fx

Hence, fx is discontinuous only at x=±1n, n∈Z-0andx=0.

Page No 9.44:

Question 18:

The value of f (0), so that the functionfx=27-2x1/3-39-3243+5x1/5x≠0 is continuous, is given by
(a) 23
(b) 6
(c) 2
(d) 4

Answer:

(c) 2

For f(x) to be continuous at x = 0, we must havelimx→0fx=f0⇒f0=limx→0fx=limx→027-2x13-39-3243+5x15⇒f0=limx→027-2x13-2713324315-243+5x15=13limx→027-2x13-2713x24315-243+5x15x=-13limx→027-2x13-2713x243+5x15-24315x=215limx→027-2x13-2713-2x243+5x15-243155x=215limx→027-2x13-271327-2x-27243+5x15-24315243+5x-243=215×13×27-2315×243-45=215×13×1272315×124345=2

Page No 9.44:

Question 19:

The value of f (0) so that the functionfx=2-256-7x1/85x+321/5-2,x ≠ 0 is continuous everywhere, is given by
(a) −1
(b) 1
(c) 26
(d) none of these

Answer:

(d) none of these

Given: fx=2-256-7x185x+3215-2

For fx to be continuous at x = 0, we must havelimx→0fx=f0⇒f0=limx→0fx=limx→02-256-7x185x+3215-2⇒f0=limx→025618-256-7x185x+3215-3215=-limx→0256-7x18-25618x5x+3215-3215x=-75limx→0256-7x18-256187x5x+3215-32155x=75limx→0256-7x18-25618256-7x-2565x+3215-32155x+32-32=75×18×256-7815×32-45=75×18×2415×27=764

Page No 9.47:

Question 38:

The points of discontinuity of the functionfx=2x,0≤x≤14-2x,1<x<522x-7,52≤x≤4isare

(a) x = 1, x=52

(b) x=52

(c) x=1,52,4

(d) x = 0, 4

Answer:

bx=52=5

If 0≤x≤1, then fx=2x.

Since fx=2x is a polynomial function, it is continuous.
Thus, fx is continuous for every 0≤x≤1.

If 1<x<52, then fx=4-2x. Since 2x is a polynomial function and 4 is a constant function, both of them are continuous. So, their difference will also be continuous.
Thus, fx is continuous for every 1<x<52.

If 52≤x≤4, then fx=2x-7. Since 2x is a polynomial function and 7 is continuous function, their difference will also be continuous.
Thus, fx is continuous for every 52≤x≤4.